I tried to draw a sphere with tikz with isometric projection or oblique projection (perpective cavalière in french).

For oblique projection you have to put a scaling factor (scal < 1) for distance on the x axes. But if I want my sphere to look like a not distorted sphere, I have to put a scaling of zero. Thus, I obtain a projection in (yz) plane. Hereafter is my tikz source code, you can easily switch between isometric and oblique projection :

Hello @altermundus I knew your web site, it's usefull. Your sphere is nice but how can I change orientation of x, y, z axes. I tried something but the sphere looks bad. I did the change on the first post.
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GerJun 7 '12 at 7:52

Here I used the "normal" oblic projection with the main plane xy. It's possible to change the factor k or the "fuite" angle, for that you need only to modify zlike this z={(-0.8cm,-0.8cm)}. It's better to modify z with an angle and a coefficient. I added an update to my answer in several minutes.
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Alain MatthesJun 7 '12 at 8:14

I do not understand. If I use a "fuite angle" of 45° z should be (-0.707,-0.707), right ? So why 0.385 and in your picture you put 3.85 istead of 0.385.
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GerJun 7 '12 at 8:50

If I put x={(-0.385cm,-0.385cm)},y={(1cm,0cm)},z={(0cm,1cm)} I get the right view. Your one is distorted with the 1.2 factor. I edit my first post in that way.
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GerJun 7 '12 at 8:51

I made a mistake when I modified my code (mm or cm ?). The correct value is -3.85 mm. If the third vector is defined by {(a,a)} then the angle is 45°. The natural projection, I think is : if the plane facing the viewer is xy , and the receding axis is z, then a point is projected like this: x'= x+0.5z.cos(a)y'= y+0.5z.sin(a) and z'=0. It's was you get 0.385 cm = 0.5 cos(45) cm. But it's possible to replace 0.5 by k and effectively in this case you get a distorsion. –
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Alain MatthesJun 7 '12 at 9:27